# How do you determine the number of possible triangles and find the measure of the three angles given a=9, c=10, mangleC=150?

Oct 31, 2017

$A = {26}^{\circ} 45 ' , B = {150}^{\circ} , C = {3}^{\circ} 15 '$

#### Explanation:

Since the given information is for a SSA triangle it is the ambiguous case. In the ambiguous case we first find the height by using the formula $h = b \sin A$.

Note that A is the given angle and its side is always a so the other side will be b .

So if $A < {90}^{\circ}$ and if

1. $h < a < b$ then then there are two solutions or two triangles.

2. $h < b < a$ then there is one solution or one triangle.

3. $a < h < b$ then there is no solution or no triangle.

If $A \ge {90}^{\circ}$ and if

1. $a > b$ then there is one solution or one triangle.

2. $a \le b$ there is no solution

$h = 9 \sin {150}^{\circ} = 4.5$, since $4.5 < 9 < 10$ we have

$h < b < a$ so we are looking for one solution. Hence,

$S \in \frac{A}{a} = \sin \frac{B}{b}$

$\sin \frac{A}{9} = \sin {150}^{\circ} / 10$

$\sin A = \frac{9 \sin {150}^{\circ}}{10}$

$A = {\sin}^{-} 1 \left(\frac{9 \sin {150}^{\circ}}{10}\right) = {26}^{\circ} 45 '$

and therefore

$C = {180}^{\circ} - {150}^{\circ} - {26}^{\circ} 45 ' = {3}^{\circ} 15 '$